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 Skip Navigation LinksMath Help > Sets, Set theory, Number systems > Vectors > Vector Product Identities

Introduction to Vector Product Identities

Let a be a vector whose components are called [a1, a2, a3], and let b and c be vectors as well.

The cross product a�b = [a2b3-a3b2, a3b1-a1b3, a1b2-a2b1].  The cross product is perpendicular to both a and b, and has a magnitude equal to |a| |b| sin θ, which is the area of a parallelogram, two sides of which are formed by vectors a and b.

The dot product a·b = |a| |b| cos θ, and in three dimensions, a·b = a1b1 + a2b2 + a3b3

The triple product (a�b)·c is a scalar representing the "signed volume" of a parallelepiped.

Lagrange's formula

a�(b�c) = (c�a)b - (b�a)c is Lagrange's formula, sometimes called the "CAB Minus BAC" identity

Proof:

Let a = [a1,a2,a3], and let b and c be defined similarly.

b�c = [b2c3-b3c2, b3c1-b1c3, b1c2-b2c1]

a�(b�c) = [a2(b1c2-b2c1)-a3(b3c1-b1c3), a3(b2c3-b3c2)-a1(b1c2-b2c1), a1(b3c1-b1c3)-a2(b2c3-b3c2)]
a�(b�c) = [a2b1c2-a2b2c1-a3b3c1+a3b1c3, a3b2c3- a3b3c2-a1b1c2+a1b2c1, a1b3c1- a1b1c3-a2b2c3+a2b3c2]

(c�a)b = [(a1c1+a2c2+a3c3)b1, (a1c1+a2c2+a3c3)b2, (a1c1+a2c2+a3c3)b3]
(c�a)b = [a1b1c1+a2b1c2+a3b1c3, a1b2c1+a2b2c2+a3b2c3, a1b3c1+a2b3c2+a3b3c3]

(b�a)c = [(a1b1+a2b2+a3b3)c1, (a1b1+a2b2+a3b3)c2, (a1b1+a2b2+a3b3)c3]
(b�a)c = [a1b1c1+a2b2c1+a3b3c1, a1b1c2+a2b2c2+a3b3c2, a1b1c3+a2b2c3+a3b3c3]

(c�a)b - (b�a)c = [a2b1c2-a2b2c1+a3b1c3-a3b3c1, a1b2c1-a1b1c2+a3b2c3-a3b3c2, a1b3c1-a1b1c3+a2b3c2-a2b2c3]

From the expansion, we see that a�(b�c) is identical to (c�a)b - (b�a)c

Some advice:

It's hard to remember what dots with what in this formula, so here's a tip.  Think about the physical representation of b�c -- it's a vector that's normal (that means perpendicular) to the plane defined by vectors b and c.  Think of a flagpole sticking straight out of a vast b-c plane.  When you cross any other vector with that flagpole, the result is perpendicular to the flagpole; in other words the result is in that vast b-c plane.  So the vector that results from a�(b�c) is in the b-c plane, which means it's a linear combination of b and c.

Now look at the formula again.  a�(b�c) = (c�a)b - (b�a)c  -- Do you see how this formula reminds you that a�(b�c) is a linear combination of b and c?  Now you need to remember just a few more things, and they should be fairly intuitive:

  1. The scalar factor for b is the dot product of the other two vectors, and likewise for c.
  2. This identity has a subtraction in it, not an addition.  That should be obvious if you consider what if b=c?  Clearly b�c = 0, so a�(b�c) has to be zero, too.
  3. Use the right-hand rule to figure out whether to subtract (c�a)b - (b�a)c or vice versa.

If you follow that advice, you should be able to discern the very similar formula for (a�b)�c -- then read on in this page to see if you were right.

Jacobi Identity

a�(b�c) + b�(c�a) + c�(a�b) = 0

Proof:

Add the "CAB Minus BAC" identities:

a�(b�c) = (c�a)b - (b�a)c
b�(c�a) = (a�b)c - (c�b)a
c�(a�b) = (b�c)a - (a�c)b 

What about (a�b)�c ?

(a�b)�c = (c�a)b - (b�c)a

Proof:

Let a = [a1,a2,a3], and let b and c be defined similarly.

a�b = [a2b3-a3b2, a3b1-a1b3, a1b2-a2b1]

(a�b)�c = [(a3b1-a1b3)c3-(a1b2-a2b1)c2, (a1b2-a2b1)c1-(a2b3-a3b2)c3, (a2b3-a3b2)c2-(a3b1-a1b3)c1]
(a�b)�c = [a3b1c3-a1b3c3-a1b2c2+a2b1c2, a1b2c1-a2b1c1-a2b3c3+a3b2c3, a2b3c2-a3b2c2-a3b1c1+a1b3c1]

This is equal to (c�a)b - (b�c)a, which you can verify using the same method as in the Lagrange formula, above.

Some advice:

In both of these identities, you have (c�a)b as a positive number, with the other vector subtracted from this one.  Look at them again:

a�(b�c) = (c�a)b - (b�a)c   -- because the resulting vector is in the b-c plane
(a�b)�c = (c�a)b - (b�c)a
   -- because the resulting vector is in the a-b plane

Call me a cab!

(OK, you're a cab!)

The other Jacobi Identity

(a�b)�c + (b�c)�a + (c�a)�b = 0

Proof:

Add the "CAB Minus BCA" identities:

(a�b)�c = (c�a)b - (b�c)a
(b�c)�a = (a�b)c - (c�a)b
(c�a)�b = (b�c)a - (a�b)c

 

Area of Triangle formed by Endpoints of Three Vectors

Three Vectors a, b, and c have a common initial point. Their endpoints form a triangle whose area is given by the magnitude of the vector:

1/2(a�b + b�c + c�a)

Proof

Triple Product of Cross Products

(a�b)�((b�c)�(c�a)) = (a�(b�c))2

Proof

Internet references

Related Pages in this website

Vector Dot Product

Vector Cross Product

Triple Product -- a·(b�c) is a scalar representing the "signed volume" of a parallelepiped

Triangle Area Using Vectors, part 1 and part 2

Triangle Area using Determinant

Matrix Math

Geometry and Trigonometry, and in particular, the Points and Lines section.

 

The webmaster and author of this Math Help site is Graeme McRae.